Theoretical Probability of Combined Events
Section: Probability | Syllabus: Cambridge Lower Secondary Checkpoint Mathematics (0862)
Sample Space Diagrams
A sample space diagram (grid) lists every possible combined outcome of two events, letting you count favourable outcomes directly.
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Question: Two fair dice are rolled. Using a 6×6 sample space diagram (36 outcomes), find P(sum = 7).
- Step 1: Pairs summing to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) - 6 outcomes
- Step 2: P(sum=7) = 6/36
- Answer: 1/6
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Question: Using the same grid, find P(sum is greater than 9).
- Step 1: Sum=10: 3 ways. Sum=11: 2 ways. Sum=12: 1 way. Total = 6 outcomes
- Step 2: P(sum>9) = 6/36
- Answer: 1/6
Sample Space Diagrams with Unequal-Sized Events
A sample space diagram isn't always a square grid - if the two events don't have the same number of outcomes, the total number of cells is (outcomes of event 1) × (outcomes of event 2), not a guessed square number.
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Question: A fair 4-sided spinner labelled 1-4 is spun and a fair 6-sided die is rolled. Using a sample space diagram, find P(the spinner and the die show the same number).
- Step 1: The spinner has 4 outcomes and the die has 6 outcomes, so the grid has 4 × 6 = 24 total outcomes, NOT 36
- Step 2: Matching pairs: (1,1), (2,2), (3,3), (4,4) - 4 outcomes (the spinner can never match a 5 or 6, since it doesn't go that high)
- Step 3: P(same number) = 4/24
- Answer: 1/6
Tree Diagrams: Independent Events
A tree diagram shows every possible outcome of successive events as branches, each labelled with its probability. Multiply along a branch (for "and"); add across different qualifying branches (for "or").
Multiply along each branch; add the highlighted branches for "exactly one head"
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Question: A biased coin has P(heads)=0.6, P(tails)=0.4. It is flipped twice. Find P(exactly one head).
- Step 1: HT = 0.6 × 0.4 = 0.24. TH = 0.4 × 0.6 = 0.24
- Step 2: P(exactly one head) = 0.24 + 0.24
- Answer: 0.48
- Check: all four final outcomes should sum to 1: 0.36 + 0.24 + 0.24 + 0.16 = 1 ✓
Tree Diagrams: Dependent Events (Without Replacement)
When events are dependent, the branch probabilities change after the first event, based on the updated totals.
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Question: A bag has 5 red and 3 blue balls (8 total). Two balls are drawn without replacement. Find P(both red).
- Step 1: 1st draw: P(red) = 5/8
- Step 2: 2nd draw (given 1st was red): 4 red left, 7 total: P(red) = 4/7
- Step 3: P(both red) = 5/8 × 4/7 = 20/56
- Answer: 5/14
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Question: Using the same bag, find P(one red and one blue), in either order.
- Step 1: P(red then blue) = 5/8 × 3/7 = 15/56
- Step 2: P(blue then red) = 3/8 × 5/7 = 15/56
- Step 3: Add both orders: 15/56 + 15/56
- Answer: 30/56 = 15/28
Finding "At Least One" Probabilities
It's often much easier to use the complement rule: P(at least one) = 1 − P(none), rather than listing every possible case.
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Question: A fair coin is flipped 3 times. Find P(at least one head).
- Step 1: P(no heads) = P(TTT) = (1/2)³ = 1/8
- Step 2: P(at least one head) = 1 − 1/8
- Answer: 7/8
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Question: A bag has 5 red and 3 blue balls. Two balls are drawn WITH replacement. Find P(at least one red).
- Step 1: P(no red, i.e. both blue) = 3/8 × 3/8 = 9/64
- Step 2: P(at least one red) = 1 − 9/64
- Answer: 55/64
Real-World Applications
Combined probabilities are used in many fields:
- Genetics: probability trees for inheritance of traits.
- Medical testing: probability of results across multiple tests.
- Manufacturing: probability of defects across multiple inspection stages.
- Sports: probability of winning multiple games in a series.
- Games of chance: card and dice games with combined outcomes.
Exam Tips
Common Mistakes
MistakeAdding probabilities along a tree diagram branch instead of multiplying
Fixmultiply probabilities ALONG a branch (for "and"); add probabilities ACROSS different branches (for "or")
MistakeUsing the same probability for both draws when events are dependent (without replacement)
Fixupdate later branch probabilities to reflect the changed totals after each draw
MistakeForgetting to include every qualifying branch/order, e.g. missing one order of "one red and one blue"
Fixidentify every branch that satisfies the condition, then add their probabilities together
MistakeListing every possible case to find "at least one" instead of using the complement rule
Fixit's often much easier to calculate P(at least one) = 1 − P(none)
MistakeMiscounting outcomes in a sample space diagram, especially near repeated sums
Fixcarefully list every cell of the grid systematically before counting favourable outcomes
MistakeAssuming a sample space diagram is always a square grid with the same total for both events
Fixthe total number of outcomes is (outcomes of event 1) × (outcomes of event 2) - check each event's own number of outcomes first
For Exams
- Use a sample space diagram for two combined events with a manageable number of outcomes.
- Use a tree diagram for successive events, especially dependent ones.
- Multiply along branches; add across different qualifying branches.
- For "at least one": calculate 1 − P(none) rather than listing every case.
- Check that all branch probabilities in a tree diagram sum to 1.
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